3

In the ring of algebraic integers $R = \Bbb Z[\xi],\ \xi^2 + \xi + 1 = 0,$ i.e. $\xi = \frac{-1+\sqrt{-3}}{2}$, I have seen three definitions for coprime ideals $(\alpha), (\beta) $:

$(1)\,\ $ Every common divisor of $ \alpha, \beta$ is a unit (invertible).

$(2)\,\ $ $ \alpha, \beta$ are not contained in a common maximal ideal.

$(3)\,\ $ The ideal sum $(\alpha)+(\beta)=(1)= R $

Are these three definitions equivalent? The inferences $(2) \Rightarrow (3) \Rightarrow (1)\,$ are straightforward (cf. comments below), but what about $(1) \Rightarrow (3)$ or $(1) \Rightarrow (2)$. These can be done proved using this Theorem, but that requires extra work - showing that $R$ is a Euclidean domain or PID. Is this extra work necessary or is there another (general) method?

Bill Dubuque
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threeautumn
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    Equivalence of $2$ and $3$ is simple. If $A$ and $B$ are those ideals and if $A, B \subset M$ then $A + B \subset M$ so if $A + B = R$ then $A$ and $B$ are not contained in a common maximal ideal. Conversely if $A$ and $B$ are not contained in a common maximal ideal then $A + B$ must be $R$ otherwise the maximal ideal containing $A + B$ contains both of them. – Rishi Jul 27 '22 at 17:41
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    The only reason 1 is equivalent to the rest is that $\mathbb{Z}[\zeta]$ is a PID/UFD; if you change to the ring $\mathbb{Z}[\sqrt{-5}]$, that equivalence no longer holds: the only elements that divide both $2$ and $1+\sqrt{-5}$ are units, but $(2)$ and $(1+\sqrt{-5})$ are both contained in $(2,1+\sqrt{-5})\neq R$. – Arturo Magidin Jul 27 '22 at 17:42
  • See also this Theorem on the equivalence with $(1)\ \ $ – Bill Dubuque Jul 27 '22 at 17:44
  • $3 \implies 1$ If $\lambda$ divides both of them then every element of $A + B$ is a multiple of $\lambda$. Now $A + B = R$ means $1\in A + B$ so some multiple of $\lambda$ is $1$ i.e. $\lambda$ is a unit. – Rishi Jul 27 '22 at 17:44
  • @Rishi, thank you. Yes, now we have (2) -> (3) -> (1), what about (1) -> (3) or (1) -> (2) – threeautumn Jul 27 '22 at 18:32
  • @BillDubuque I feel using that thm requires to prove that $Z[\xi]$ is Euclidean domain or PID which requires extra work. – threeautumn Jul 27 '22 at 18:35

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